Techniques of Integration: math lesson, College

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Math lessons College : Techniques of Integration — Zyro the alien explorer of Planète Maths

Some integrals fall to a quick guess; most do not. In this chapter you build a toolbox of six methods (parts, trigonometric integrals, trigonometric substitution, partial fractions, improper integrals, and numerical rules) and, just as important, you learn how to pick the right tool in a few seconds.

Sommaire

1. Why you need a toolbox

The Fundamental Theorem of Calculus says that an integral is easy once you know an antiderivative. The difficulty is finding one. Differentiation follows a short list of rules and always works; integration has no single rule, so we rewrite the integrand until it matches something familiar. Each technique below is a different way of rewriting.

Keep the basic forms ready: \( \int x^n\,dx = \dfrac{x^{n+1}}{n+1}+C \) for \( n\neq -1 \), \( \int \dfrac{dx}{x} = \ln|x|+C \), \( \int e^{kx}dx=\dfrac{e^{kx}}{k}+C \), and \( \int \cos x\,dx=\sin x+C \). The ordinary \(u\)-substitution (for example \( \int x e^{x^2}dx=\tfrac12 e^{x^2}+C \)) is always the first thing to test.

2. Integration by parts

The product rule \( (uv)' = u'v+uv' \) integrates to a formula that trades one integral for another, hopefully simpler, one.

Integration by parts \[ \int u\,dv = uv-\int v\,du, \qquad \int_a^b u\,v'\,dx = \Big[uv\Big]_a^b-\int_a^b u'\,v\,dx. \]

Method

Split the integrand into a part \(u\) that gets simpler when differentiated and a part \(dv\) you can integrate.
Write \(du\) and \(v\).
Apply the formula and finish the new integral.

Good choices for \(u\), in rough order of priority: a logarithm, an inverse trig function, a polynomial, then a trig or exponential factor.

Example 1: polynomial times exponential Compute \( \int_0^1 x e^{2x}\,dx \). Take \(u=x\), \(dv=e^{2x}dx\), so \(du=dx\) and \(v=\tfrac12 e^{2x}\). Then
\[ \int_0^1 xe^{2x}dx=\Big[\tfrac{x}{2}e^{2x}\Big]_0^1-\tfrac12\int_0^1 e^{2x}dx=\tfrac{e^2}{2}-\tfrac{e^2-1}{4}=\dfrac{e^2+1}{4}\approx 2.097. \]

Example 2: a logarithm alone For \( \int \ln x\,dx \) take \(u=\ln x\), \(dv=dx\). Then \(du=\dfrac{dx}{x}\), \(v=x\) and \( \int\ln x\,dx = x\ln x-\int 1\,dx = x\ln x-x+C \).

Example 3: going around in a circle For \( I=\int e^x\cos x\,dx \), parts twice gives \( I=e^x\cos x+e^x\sin x-I \). The original integral reappears, so solve for it: \( I=\dfrac{e^x(\cos x+\sin x)}{2}+C \).

3. Trigonometric integrals

Integrals made of powers of sine and cosine are handled by the identities \( \sin^2x+\cos^2x=1 \), \( \sin^2x=\dfrac{1-\cos 2x}{2} \) and \( \cos^2x=\dfrac{1+\cos 2x}{2} \).

Method for \( \int \sin^m x\cos^n x\,dx \)

If one exponent is odd, peel off one factor of that function, convert the rest with \( \sin^2+\cos^2=1 \), and substitute for the other function.
If both exponents are even, lower the powers with the half-angle identities.
For \( \tan \) and \( \sec \), use \( \tan^2x=\sec^2x-1 \) and \( \dfrac{d}{dx}\tan x=\sec^2x \).

Example 4: odd power \( \int \sin^3x\cos^2x\,dx=\int(1-\cos^2x)\cos^2x\,\sin x\,dx \). With \(u=\cos x\), \(du=-\sin x\,dx\): \( -\int(u^2-u^4)du=-\dfrac{\cos^3x}{3}+\dfrac{\cos^5x}{5}+C \).

Example 5: even power \( \int\sin^2x\,dx=\int\dfrac{1-\cos2x}{2}dx=\dfrac{x}{2}-\dfrac{\sin 2x}{4}+C \).

4. Trigonometric substitution

When the integrand contains \( \sqrt{a^2-x^2} \), \( \sqrt{a^2+x^2} \) or \( \sqrt{x^2-a^2} \), a trigonometric identity can remove the root.

Expression in the integrand	Substitution	Identity used
\( \sqrt{a^2-x^2} \)	\( x=a\sin\theta \)	\( 1-\sin^2\theta=\cos^2\theta \)
\( \sqrt{a^2+x^2} \)	\( x=a\tan\theta \)	\( 1+\tan^2\theta=\sec^2\theta \)
\( \sqrt{x^2-a^2} \)	\( x=a\sec\theta \)	\( \sec^2\theta-1=\tan^2\theta \)

Example 6 Compute \( \int\sqrt{9-x^2}\,dx \). Let \(x=3\sin\theta\), \(dx=3\cos\theta\,d\theta\), \( \sqrt{9-x^2}=3\cos\theta \). Then \( \int 9\cos^2\theta\,d\theta=\dfrac92(\theta+\sin\theta\cos\theta)+C \). Returning to \(x\) with \( \sin\theta=\tfrac x3 \) and \( \cos\theta=\tfrac{\sqrt{9-x^2}}{3} \):
\[ \int\sqrt{9-x^2}\,dx=\dfrac92\arcsin\dfrac x3+\dfrac x2\sqrt{9-x^2}+C. \]

Example 7 Compute \( \int\dfrac{dx}{(x^2+4)^{3/2}} \). Let \(x=2\tan\theta\), \(dx=2\sec^2\theta\,d\theta\), \( (x^2+4)^{3/2}=8\sec^3\theta \). The integral becomes \( \int\dfrac{\cos\theta}{4}d\theta=\dfrac{\sin\theta}{4}+C=\dfrac{x}{4\sqrt{x^2+4}}+C \), read off a right triangle with legs \(x\) and \(2\).

Watch out Always convert back to \(x\) (or change the limits to \(\theta\)). A triangle sketch is the safest way to do it, and it avoids sign mistakes.

5. Partial fractions

A rational function \( \dfrac{P(x)}{Q(x)} \) with \( \deg P<\deg Q \) can be split into simple pieces. If \( \deg P\ge\deg Q \), divide first.

Method

Factor \(Q(x)\) completely.
Each linear factor \((x-r)\) gives \( \dfrac{A}{x-r} \); a repeated factor \((x-r)^2\) gives \( \dfrac{A}{x-r}+\dfrac{B}{(x-r)^2} \); an irreducible quadratic gives \( \dfrac{Ax+B}{x^2+px+q} \).
Clear denominators and find the constants (plug in roots, or compare coefficients).
Integrate each piece.

Example 8 \( \dfrac{5x+7}{(x-1)(x+3)}=\dfrac{A}{x-1}+\dfrac{B}{x+3} \). Then \( 5x+7=A(x+3)+B(x-1) \). At \(x=1\): \(12=4A\), so \(A=3\). At \(x=-3\): \(-8=-4B\), so \(B=2\). Therefore \( \int\dfrac{5x+7}{(x-1)(x+3)}dx=3\ln|x-1|+2\ln|x+3|+C \).

Example 9: repeated factor \( \dfrac{3x+1}{x^2(x+1)}=\dfrac Ax+\dfrac B{x^2}+\dfrac C{x+1} \) gives \(B=1\), \(C=-2\), \(A=2\). So \( \int=2\ln|x|-\dfrac1x-2\ln|x+1|+C \).

6. Improper integrals

An integral is improper if an endpoint is infinite or the integrand blows up on the interval. We replace the trouble spot by a limit.

Definition \( \displaystyle\int_a^{\infty}f(x)\,dx=\lim_{b\to\infty}\int_a^bf(x)\,dx \). If the limit is a finite number, the integral converges; otherwise it diverges. Similarly, if \(f\) blows up at \(a\), use \( \lim_{t\to a^+}\int_t^b f\,dx \).

The p-test \( \displaystyle\int_1^{\infty}\dfrac{dx}{x^p} \) converges (to \( \dfrac1{p-1} \)) when \(p\gt1\) and diverges when \(p\le1\). Also \( \displaystyle\int_0^1\dfrac{dx}{x^p} \) converges (to \( \dfrac1{1-p} \)) when \(p\lt1\).

Example 10 \( \int_0^{\infty}e^{-2x}dx=\lim_{b\to\infty}\Big[-\tfrac12e^{-2x}\Big]_0^b=\tfrac12 \). And \( \int_0^1\dfrac{dx}{\sqrt x}=\lim_{t\to0^+}\big[2\sqrt x\big]_t^1=2 \).

When you cannot find an antiderivative, compare: if \(0\le f\le g\) and \( \int g \) converges, then \( \int f \) converges; if \( \int f \) diverges, so does \( \int g \).

7. Numerical integration

Some antiderivatives (such as that of \( e^{-x^2} \)) cannot be written with elementary functions. We then approximate with \(n\) subintervals of width \( h=\dfrac{b-a}{n} \) and \( x_i=a+ih \).

Trapezoid and Simpson’s rules \[ T_n=\dfrac h2\big[f_0+2f_1+\cdots+2f_{n-1}+f_n\big],\qquad S_n=\dfrac h3\big[f_0+4f_1+2f_2+4f_3+\cdots+4f_{n-1}+f_n\big], \] where \( f_i=f(x_i) \) and \(n\) must be even for Simpson’s rule. The trapezoid error is at most \( \dfrac{K(b-a)^3}{12n^2} \) with \( K\ge|f''| \).

Example 11 Estimate \( \ln 3=\int_1^3\dfrac{dx}{x} \) with \(n=4\), \(h=0.5\). The values are \(1,\ \tfrac23,\ \tfrac12,\ \tfrac25,\ \tfrac13\). Then \( T_4=0.25\big[1+2(\tfrac23+\tfrac12+\tfrac25)+\tfrac13\big]\approx1.1167 \) and \( S_4=\dfrac{0.5}{3}\big[1+\tfrac83+1+\tfrac85+\tfrac13\big]=1.1 \). The true value is \( \ln 3\approx1.0986 \), so Simpson’s rule is much closer.

8. Choosing a technique

Run through this checklist in order and stop at the first match.

If you see…	Try…
A form that matches a basic antiderivative after a simple change of variable	\(u\)-substitution
A product such as \( x e^x \), \( x\sin x \), \( \ln x \) or \( x^n\ln x \)	Integration by parts
Powers of \( \sin \), \( \cos \), \( \tan \), \( \sec \)	Trigonometric identities
\( \sqrt{a^2\pm x^2} \) or \( \sqrt{x^2-a^2} \)	Trigonometric substitution
A rational function with a factorable denominator	Partial fractions
An infinite limit or a vertical asymptote	Replace it by a limit (improper integral)
No elementary antiderivative, or only data	Trapezoid or Simpson

Zyro’s tip On my home planet we say: simplify first, substitute second, split third. Many “hard” integrals turn easy after a quick algebra step or an obvious \(u\)-substitution.

Classic mistakes Forgetting the minus sign in \( uv-\int v\,du \); forgetting to divide first when the numerator’s degree is too high; and writing \( \int_1^{\infty}\dfrac{dx}{x}=0 \) because “1/x goes to 0” (it diverges).

Key takeaways

Parts: \( \int u\,dv=uv-\int v\,du \); choose \(u\) as the factor that simplifies when differentiated (logs first).
Odd power of sine or cosine: peel one factor and substitute; even powers: half-angle identities.
\( \sqrt{a^2-x^2} \to x=a\sin\theta \); \( \sqrt{a^2+x^2} \to x=a\tan\theta \); \( \sqrt{x^2-a^2} \to x=a\sec\theta \).
Partial fractions: factor, decompose, solve for constants, integrate; divide first if the degree is too big.
Improper integrals are limits; the p-test: \( \int_1^\infty x^{-p}dx \) converges exactly when \(p\gt1\).
Trapezoid and Simpson’s rules approximate integrals; Simpson’s needs an even \(n\) and is usually far more accurate.